millimols NH3 initial = 250 x 0.05 = 12.5
mm NH4Cl initial = 250 x 0.15 = 37.5
add 0.01 mol = 10 mmols HCl.
.......NH3 + HCl ==> NH4Cl
I......12.5..0.0.....37.5
add..........10..........
C.....-10...-10......+10
E......2.5....0......47.5
Substitute the E line into Henderson-Hasselbalch equation and solve for pH.
b is done the same way.
Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL of each of the following buffered solutions:
a. .050 M NH3/.15 M NH4Cl
b. .5 M NH3/ 1.5 M NH4Cl
1 answer
1 answer